Has topological sine a primitive? I need to know, for one summation of series, if the topological sine define as
\begin{align}
f:\mathbb{R}&\rightarrow\mathbb{R}\newline
x&\mapsto \begin{cases} \sin\frac{1}{x}&\text{ if }x\neq 0\newline
0&\text{ if }x=0\end{cases} 
\end{align}
has an antiderivative or primitive.
 A: Not in terms of elementary functions:
$$\int \sin\left(\frac{1}{x}\right)dx = x \sin\left(\frac{1}{x}\right) - \operatorname{Ci}\left(\frac{1}{x}\right) + C$$
where $\operatorname{Ci}(x)$ is the cosine integral function: $\operatorname{Ci}(x) = -\int_{x}^{\infty} \frac{\cos(t)}{t}dt$.
you can see this by substituting $u = \frac{1}{x}$, then $dx = -\frac{1}{u^2}$, and you can integrate $-\int \frac{\sin(u)}{u^2}du$ by parts
A: It does: $F(x) = x \sin(1/x) - \text{Ci}(1/x)$ for $x > 0$, $x \sin(1/x) - \text{Ci}(-1/x)$ for $x < 0$, $0$ for $x=0$, where $\text{Ci}$ is the "cosine integral" function.  Note that since $\text{Ci}(t) = \frac{\sin(t)}{t} + O(1/t^2)$ as $t \to +\infty$, $F(x) = O(x^2)$ as $x \to 0$ so $F'(0) = 0$.
A: An alternate way to say it:  $f$ is Riemann integrable (on bounded intervals) so the function $F$ defined by $F(x) = \int_0^x f(t)\,dt$  has the property $F'(x) = f(x)$  at every point where $f$ is continuous.  That is, at every point except $0$.  Additional study would be required to see whether $F'(0)$ exists or not.  That additional study is presumably the same as showing the integral defining Ci converges.
